The ka of propanoic acid (c2h5cooh) is 1.34 × 10-5. calculate the ph of the solution and the concentrations of c2h5cooh and c2h5coo– in a 0.193 m propanoic acid solution at equilibrium.

Chemical reaction: C₂H₅COOH ⇄ C₂H₅COO⁻ + H⁺.
Ka(C₂H₅COOH) = 1,34·10⁻⁵.
c(C₂H₅COOH) = 0,193 M = 0,193 mol/L.
[C₂H₅COO⁻] = [H⁺] = x.
Ka = [C₂H₅COO⁻] · [H⁺] / [C₂H₅COOH].
1,34·10⁻⁵ = x² / (0,193 M - x).
Solve quadratic equation: [C₂H₅COO⁻] = [H⁺] = 0,00158 M.
[C₂H₅COOH] = 0,193 mol/L - 0,0158 M = 0,1772 M.
pH = -log[H⁺].
pH = -log(0,00158 M) = 2,8.

I'm not sure but i will try to find out